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8 tháng 5 2022

đk x > 0 

\(\dfrac{A}{B}=\dfrac{\dfrac{x+2\sqrt{x}}{x}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}+2}{\sqrt{x}+1}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{7}{4}< 0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}+4-7\sqrt{x}}{4\sqrt{x}}< 0\Leftrightarrow\dfrac{-3\sqrt{x}+4}{4\sqrt{x}}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3\sqrt{x}+4\ne0\\-3\sqrt{x}+4< 0\\4\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{16}{9}\\x< \dfrac{16}{9}\\x\ne0\end{matrix}\right.\)

a: ĐKXĐ: x>0; x<>4

b: \(A=\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{4-x}:\dfrac{2\sqrt{x}-\sqrt{x}-3}{2\sqrt{x}-x}\)

\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{4-x}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4x}{\sqrt{x}-3}\)

c: \(A-1=\dfrac{4x-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9 và x<>4

7 tháng 5 2022

mik cần gấp ạ^^

 

26 tháng 5 2021

\(A=\dfrac{-\left(\sqrt{x}+1\right)\left(2+\sqrt{x}\right)-2\sqrt{x}\left(2-\sqrt{x}\right)+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)

\(A=\dfrac{-3\sqrt{x}-x-2-4\sqrt{x}+2x+5\sqrt{x}+2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-x-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)^3}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)\sqrt{x}\left(3-\sqrt{x}\right)}=\dfrac{-\left(\sqrt{x}+2\right)^2}{\left(2-\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)

 

26 tháng 5 2021

Mình sửa đầu bài

31 tháng 10 2021

a) ĐKXĐ: \(x>0,x\ne1\)

\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}=\dfrac{2\sqrt{x}}{x+\sqrt{x}+1}\)

18 tháng 9 2023

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)

`b,` Tớ tính mãi ko ra, xl cậu nha=')

 

 

 

19 tháng 9 2023

b) Xét hiệu:

\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)

\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)

\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)

Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\) 

\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)

Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)

(giúp cậu nó nha) 

4 tháng 9 2023

Ta có:

\(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) (ĐK: \(x\ne4;x\ge0\)

\(B=\dfrac{x}{\left(\sqrt{x}\right)^2-2^2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(B=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(B=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(\Rightarrow P=\dfrac{A}{B}=\dfrac{\dfrac{\sqrt{x}+2}{\sqrt{x}}}{\dfrac{\sqrt{x}}{\sqrt{x}-2}}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\sqrt{x}}=\dfrac{x-4}{x}\) (ĐK: \(x\ne0\)

Theo đề ta có:

\(P\cdot x\le10\sqrt{x}-29-\sqrt{x}+25\) (ĐK: \(x\ge0\)

\(\Leftrightarrow\dfrac{x-4}{x}\cdot x\le9\sqrt{x}-4\)

\(\Leftrightarrow x-4\le9\sqrt{x}-4\)

\(\Leftrightarrow x-9\sqrt{x}\le0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-9\right)\le0\)

Mà: \(\sqrt{x}\ge0\)

\(\Leftrightarrow\sqrt{x}-9\le0\)

\(\Leftrightarrow\sqrt{x}\le9\)

\(\Leftrightarrow x\le81\)

Kết hợp với đk:

\(0\le x\le81\)

Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+8\sqrt{x}-5-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\)

\(\Leftrightarrow A\le\dfrac{2}{3}\)